Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into sets of k consecutive numbers
Return True if its possibleotherwise return False.
Example 1:
Input: nums = [1,2,3,3,4,4,5,6], k = 4
Output: true
Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
Example 2:
Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3
Output: true
Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
Example 3:
Input: nums = [3,3,2,2,1,1], k = 3
Output: true
Example 4:
Input: nums = [1,2,3,4], k = 3
Output: false
Explanation: Each array should be divided in subarrays of size 3.
Constraints:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
1 <= k <= nums.length
class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
if (nums == null || nums.length % k != 0)
return false;
// Frequency map
Map<Integer, Integer> map = new HashMap<>();
for (int num : nums)
map.put(num, map.getOrDefault(num, 0) + 1);
for (int num : nums) {
if (!map.containsKey(num))
continue;
// find the smallest value in current streak
while (map.containsKey(num))
num--;
// Taking all the possible streaks of size k
int start = num + 1, next = -1;
// next holds the start of another streak in this continuous chain
while (true) {
for (int i = 0; i < k; i++) {
// a broken streak, return false.
if (!map.containsKey(start + i))
return false;
int frequency = map.get(start + i);
if (frequency == 1)
map.remove(start + i);
else {
map.put(start + i, frequency - 1);
if (next == -1)
next = start + i;
}
}
// end of current streak
if (next == -1)
break;
start = next;
next = -1;
}
}
return true;
}
}